rigid body equilibrium five

(1)

S2013abn Rigid Body Equilibrium 1

Lecture 5 Architectural Structures ARCH 331

five

rigid body

equilibrium

lecture

A

RCHITECTURAL

S

TRUCTURES

:

F

ORM,

B

EHAVIOR, AND

D

ESIGN

A

RCH 331

HÜDAVERDİ TOZAN

(2)

Equilibrium

• rigid body

– doesn’t deform

– coplanar force systems

• static:

A

C

B

0 





_x

x

F

R

0 





_y

y

F

R

0 





M

(3)

S2013abn Architectural Structures

ARCH 331

Free Body Diagram

• FBD (sketch)

• tool to see all forces on a body or a

point including

– external forces

– weights

– force reactions

– external moments

– moment reactions

– internal forces

Rigid Body Equilibrium 3 Lecture 5

(4)

Free Body Diagram

• determine body

• FREE it from:

– ground

– supports & connections

• draw all external forces

acting ON the body

– reactions

– applied forces

– gravity

m



g

+ weight

100 lb

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ARCH 331

Free Body Diagram

• sketch FBD with relevant geometry

• resolve each force into components

– known & unknown angles –

name

them

– known & unknown forces –

name

them

– known & unknown moments –

name

them

• are any forces related to other forces?

• for the unknowns

• write only as many equilibrium equations as

needed

• solve up to 3 equations

(6)

Free Body Diagram

• solve equations

– most times 1 unknown easily solved

– plug into other equation(s)

• common to have unknowns of

– force magnitudes

– force angles

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ARCH 331

Reactions on Rigid Bodies

• result of applying force

• unknown size

• connection or support type

– known direction

– related to motion prevented

no vertical motion

no translation

no rotation

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ARCH 331

Supports and Connections

(10)

FBD Example

• 500 lb known

• pin – A

_x

, A

_y

• smooth surface –

B at 4:3

• 3 equations

• sum moments at

– A?

– B?

(B

_x

)

(11)

S2013abn Rigid Body Equilibrium 11

Lecture 5

Architectural Structures ARCH 331

Moment Equations

• sum moments at intersection where the

most forces intersect

• multiple moment equations may not be

useful

• combos:

0 F

_x





0 F

_y





0 M

1 



0 F





0 M

₁





0 M

2 



0 M

₁





0 M

₂





0 M

3 



(12)

Recognizing Reactions

F

_F

unknowns

3 weight

m



g

unknowns

3

(13)

S2013abn Architectural Structures ARCH 331

Recognizing Reactions

unknowns

3 unknowns for

2 bodies

6 unknowns

2 m



g

weight

F

₁

F

₂

weight

F

₁

F

₂

m



g

not independent

(14)

Constraints

• completely constrained

– doesn’t move

– may not be statically determinate

• improperly or partially constrained

– has



unknowns

(15)

S2013abn Architectural Structures ARCH 331

Constraints

• overconstrained

– won’t move

– can’t be solved with statics

– statically indeterminate to n

th

_degree

A

C

B

200 lb-ft

60 lb

55 

A

5’

9’

C

B

200 lb-ft

55 

60 lb

A

_x

A

_y

_C

y

M

_RA

(16)

Partial Constraints

100 N

1 m

0.75 m

30 

A

B

100 N

1 m

0.75 m

30 

A

B

A

B

W

500 mm

200 mm

B

W

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ARCH 331

Cable Reactions

• equilibrium:

– more reactions (4) than equations

– but, we have slope relationships

– x component the same everywhere

A

C

45 kN

4 m

2 m

B

6 m

45 kN

(18)

Two Force Rigid Bodies

• equilibrium:

– forces in line, equal and opposite

A

B

C

A

F

2

B

F

1

d

A

F

2

B

F

1

d

A

F

2

B

F

1

a

(no)

(19)

ARCH 331

Three Force Rigid Bodies

• equilibrium:

– concurrent or parallel forces

A

B

C

F

1

F

2

A

B

C

F

3

F

2

A

F

1

B

C

F

3

d

1

d

2

F

2

A

F

1

B

C

F

3

a

(no)

beams!

(20)

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ARCH 331

Distributed Loads

(22)

Beam Supports

• statically determinate

• statically indeterminate

L

simply supported

(most common)

overhang

cantilever

L

continuous

(most common case when L

=L

)

L

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ARCH 331

Equivalent Force Systems

• replace forces by resultant

• place resultant where M = 0

• using calculus and area centroids

dx

w(x)

x

L

lo ad in g

L

0 wdx

dA

A

W











dx

y

x

el

x

(24)

Load Areas

• area is width x “height” of load

• w is load per unit length

• W is total load

x

x/2

W

x/2

x

2x/3

W/2

x/3

x

x/2

W

x/6 x/3

W/2

0 W

x

w





_w

w

2

2 W

x

w





w

2w

(25)

ARCH 331

Method of Sections

• relies on internal forces being in

equilibrium on a section

• cut to expose 3 or less members

• coplanar forces





M = 0 too

A

_B

C

P

F

E

D

P

.

A

B

y

AC

AB

(26)

Method of Sections

• joints on or off the section are good to

sum moments

• quick for few members

• not always obvious where to cut or sum

A

_B

C

P

F

E

D

P

.

A

B

y

AC

AB

B

.